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A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure.  The coefficient of friction, between the particle and the rough track equals µ.  The particle is released, from rest, from the point P and it comes to rest at a point R.  The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction µ and the distance x(=QR), are, respectively close to :
Option: 1  0.2 and 6.5 m  
Option: 3  0.2 and 3.5 m  
Option: 4   0.29 and 6.5 m
 

Work done by friction at QR = μmgx

In triangle, sin 30° = 1/2 = 2/PQ

PQ = 4 m

Work done by friction at PQ = μmg × cos 30° × 4 = μmg × √3/2 × 4 = 2√3μmg

Since work done by friction on parts PQ and QR are equal,

μmgx = 2√3μmg

x = 2√3 ≅ 3.5 m

Applying work energy theorem from P to R

 

decrease in P.E.=P.E.= loss of energy due to friction in PQPQ and QR

\\ m g h=(\mu m g \cos \theta) P Q+\mu m g \times Q R\\ h=\mu \cos \theta \times P Q+\mu m g \times Q R\\ h=\mu \cos \theta \times P Q+\mu \times Q R =\mu \cos 30^{\circ} \times 4+\mu \times 2 \sqrt{3} =\mu\left(4 \times \frac{\sqrt{3}}{2}+2 \sqrt{3}\right)\\ h=\mu \times 4 \sqrt{3}\\ \mu=\frac{2}{4 \sqrt{3}}=\frac{1}{2 \sqrt{3}}=0.29where h=2(given)

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Ritika Jonwal

 A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed \nu in the x-y plane as shown in the figure : Which of the following statements is false for the angular momentum  \vec{L} about the
origin?
Option: 1 \vec{L}= -\frac{mv}{\sqrt{2}}R\hat{k}when the particle is moving from A to B.

Option: 2 \vec{L}= mv\left [ \frac{R}{\sqrt{2}}-a \right ]\hat{k}when the particle is moving from C to D.

Option: 3 \vec{L}= mv\left [ \frac{R}{\sqrt{2}}+a \right ]\hat{k}when the particle is moving from B to C.

Option: 4 L=\frac{R}{\sqrt{2}} m v(-k)when the particle is moving from D to A.
 

\\ In\ option\ (a), co-ordinates \ of\ A are \left(\frac{R}{\sqrt{2}}, \frac{R}{\sqrt{2}}\right) \\ \therefore \vec{r}=\left(\frac{R}{\sqrt{2}} \hat{i}+\frac{R}{\sqrt{2}} \hat{j}\right) and \vec{v}=v \hat{i}\\ \vec{L} m(\vec{r} \times \vec{v})=m\left(\frac{R}{\sqrt{2}} \hat{i}+\frac{R}{\sqrt{2}} \hat{j}\right) \times v \hat{i}\\ \vec{L}=-\frac{m R}{\sqrt{2}} v \hat{k}

\\ in\ option \ (b)\ it \ moves\ from \ C \ to\ D\\ L=\left(\frac{R}{\sqrt{2}}+a\right) m v(\hat{k})

so option b is correct option

\\ in \ option\ (c), \ For\ B$ \ to\ $C,$ \\ we have $L=\left(\frac{R}{\sqrt{2}}+a\right) m v(\hat{k})

\\ in \ option \ (d), \ When \ a \ particle\ is \ moving\ from \ D \ to\ A\\ L=\frac{R}{\sqrt{2}} m v(-k)

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Ritika Jonwal

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A body of mass m=10−2 kg is moving in a medium and experiences a frictional forceF= -kv^{2}  Its initial speed is v_{0}= 10ms^{-1}.  If, after 10 s, its energy is \frac{1}{8}  mv{_{0}}^{2}  the value of k will be :  
Option: 1 10−3 kg m−1  
Option: 2  10−3 kg s−1  
Option: 3  10−4 kg m−1  
Option: 4 10−1 kg m−1 s−1  
 

As we learnt in 

\frac{1}{2}mv{_{f}}^{2}=\frac{1}{8}mv{_{o}}^{2}

v_{f}=\frac{Vo}{2}=5 m \slash s

\left ( 10^{-2} \right )\frac{dv}{dt}=-KV^{2}

\int_{10}^{5}=-100k\int_{0}^{10}dt\Rightarrow \frac{1}{5}-\frac{1}{10}=100 k (1^{\circ})

K=10^{-4}

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vishal kumar

Consider a uniform rod of massM=4m and length l pivoted about its centre. A mass m moving with velocity v making angle \theta =\frac{\pi }{4} to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod - mass system just after the collision is :   
Option: 1 \frac{4}{7}\frac{v}{l}

Option: 2 \frac{3\sqrt{2}}{7}\frac{v}{l}

Option: 3 \frac{3}{7\sqrt{2}}\frac{v}{l}  

Option: 4 \frac{3}{7}\frac{v}{l}

 

Let us conserve angular momentum about O:-

           So, L_i=\left ( \frac{mv}{\sqrt2} \right )\times \frac{l}{2}, where \left ( \frac{mv}{\sqrt2} \right ) is linear momentum and \left ( \frac{l}{2} \right ) is the distance from centre O.

Now, L_f=I\omega

Here, I=\frac{4ml^2}{12}+{m\left (\frac{l}{2} \right )^2}=\frac{7ml^2}{12}

So, L_i=L_f\Rightarrow \omega=\frac{6v}{7\sqrt2 l}= \frac{3\sqrt2v}{7 l}

So the correct graph is given in option 2.

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vishal kumar

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A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l. The other end is fixed. The system is given an angular speed \omega about the fixed end of the spring such that it rotates in a circle in gravity-free space. Then the stretch in the spring is :     
Option: 1 \frac{ml\omega ^2}{k-\omega m}
 
Option: 2 \frac{ml\omega ^2}{k-m\omega ^2}

Option: 3 \frac{ml\omega ^2}{k+m\omega ^2}  

Option: 4 \frac{ml\omega ^2}{k+m\omega }
 

 

 

 

 

As natural lentgh=l

Let elongation=x

Mass m is moving with angular velocity \omega in a radius r

where r=l+x

Due to elongation x spring force is given by F_s=Kx

And F_C=m\omega ^2r=m\omega ^2(l+x)

as F_C=F_s

So

 Kx =m\omega ^2(l+x)\\ \Rightarrow x=\frac{m\omega ^2l}{K-m\omega ^2}

So the correct option is 2.

 

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vishal kumar


The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plate) of mass 4kg. (The coordinates of the same are shown in the figure) are :
Option: 1 (1.25m,1.50.m)
Option: 2 (0.75m,0.75m)
Option: 3 (0.75m,1.75m)
Option: 4 (1m,1.75m)
 

 

 

 

 

The co-ordinate of O1 is (0.5,1),  O2 is (1, 2.5)

\\x_{cm}=\frac{m\times0.5+m\times1}{2m}=0.75\\ y_{cm}=\frac{m\times+m\times\frac{5}{2}}{2m}=1.75

So the correct option is 3.

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vishal kumar

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A spring mass system (mass m, spring constant k and natural length l ) rests in equilibrium on a horizontal disc.The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotates about its axis with an angular velocity \omega, (k> > m\omega ^{2}) the relative change in the length of the spring is best given by the option :
Option: 1 \frac{m\omega ^{2}}{3k}      
Option: 2  \frac{m\omega ^{2}}{k}       
Option: 3  \sqrt{\frac{2}{3}}\left ( \frac{m\omega ^{2}}{k} \right )      
Option: 4 \frac{2m\omega ^{2}}{k}
 

As natural lentgh=l0

Let elongation=x

Mass m is moving with angular velocity \omega in a radius r

where r=l_{0}+x

Due to elongation x spring force is given by F_{s}=Kx

And F_{C}=m\omega ^{2}r=m\omega ^{2}(l_{0}+x)

as F_{C}=F_{s}

Kx=m\omega ^{2}(l_{0}+x)

\Rightarrow x=\frac{m\omega ^{2}l_{0}}{K-m\omega ^{2}}

Using k> > m\omega ^{2}

So, \frac{x}{l_{0}} is equal to \frac{m\omega ^{2}}{k}

Hence the correct option is (2).

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avinash.dongre

Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at three corners of a right triangle of sides 4.0 cm, 3.0 cm, and 5.0 cm as shown in the figure. The centre of mass of the system is at a point:
 

Option: 1 2.0\; cm right and 0.9\; cm above 1\: kg mass
   
Option: 2 0.9\; cm right and 2.0\; cm above 1\: kg mass

Option: 3 0.6\; cm right and 2.0\; cm above 1\: kg mass  

Option: 4 1.5\; cm right and 1.2\; cm above 1\: kg mass
 

 

 

let m1=1 kg, m2=1.5 kg and m3=2.5 kg

x1=0, x2=3, x3=0 and y1=0, y2=0, y3=4

x_{com}=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} and y_{com}=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}

Let point A be origin and mass m1=1.0 kg be at origin.

So, x_{com}=\frac{1\times0+1.5\times3+2.5\times0}{1+1.5+2.5}=\frac{4.5}{5}=0.9

and y_{com}=\frac{1\times0+1.5\times0+2.5\times4}{1+1.5+2.5}=\frac{10}{5}=2

so centre of mass of the system is at (0.9,2).

So from the figure we can say that the 0.9 cm right and 2 cm above the 1 kg mass. 

So option (2) is correct.

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Ritika Jonwal

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As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a flywheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance h, the angular speed of the wheel will be:-
 
Option: 1 r\sqrt{\frac{3}{2gh}}
 
Option: 2 r\sqrt{\frac{3}{4gh}}

Option: 3 \frac{1}{r}\sqrt{\frac{4gh}{3}}  

Option: 4 \frac{1}{r}\sqrt{\frac{2gh}{3}}
 

 

 

 

Conservation Of angular momentum -    

\\mg-T=ma\\T\times r=I\alpha\\T=\frac{mr^2}{2}\times\frac{a}{r}\times\frac{1}{r}\\T=\frac{ma}{2}\\mg=\frac{3ma}{2}\\a=\frac{2g}{3}\\Also,\ v=\sqrt{2as}=\sqrt{\frac{4gh}{3}}\\also, v=\omega r\\ \omega=\frac{v}{r}

\Rightarrow \sqrt{\frac{4gh}{3}}\times\frac{1}{r}=\frac{1}{r}\sqrt{\frac{4gh}{3}}

Hence option (3) is correct.

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Ritika Jonwal

The radius of gyration of a uniform rod of length l, about an axis passing through a point \frac{l}{4} away from the centre of the rod and perpendicular to it is:-  
Option: 1 \frac{1}{8}l

 
Option: 2 \sqrt{\frac{7}{48}l}

Option: 3 \sqrt{\frac{3}{8}l}  

Option: 4 \frac{1}{4}l
 

 

 

                  

Moment of inertia of rod about an axis perpendicular through COM

I_{COM}=\frac{Ml^2}{12}

And I_N=I_{COM}+M(\frac{l}{4})^2=\frac{Ml^2}{12}+M(\frac{l^2}{16})=\frac{7Ml^2}{48}

Radius of Gyration

\\ K=\sqrt{\frac{I_N}{M}}=\sqrt{\frac{7Ml^2}{48M}}=\frac{l}{4}*\sqrt{\frac{7}{3}} \\ = l\sqrt{\frac{7}{48}}

 

So the option (2) is correct.

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Ritika Jonwal

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